Question

The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value

Answer 1

The potential difference between points a and b will reach one-half of its initial value after one time constant in an RL circuit.

Step-by-step explanation:

The time it takes for the potential difference between points a and b to reach one-half of its initial value after the current is suddenly turned off can be determined using the concept of the time constant in an RL circuit. The time constant, denoted by τ, is given by the equation τ = L/R, where L is the inductance and R is the resistance in the circuit.

After one time constant, the potential difference decreases to approximately 0.368 of its initial value. So, it will take approximately one time constant for the potential difference to reach one-half of its initial value.

Example: If the time constant in a given RL circuit is 2 seconds, then it will take 2 seconds for the potential difference between points a and b to reach one-half of its initial value.

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

Step-by-step explanation:

From the question we are told that

The original voltage is
`V_o`

The new voltage is
`V  =(V_o)/(2)`

The capacitance is
`C = 150\ nF = 150 *10^(-9) \  F`

The first resistance is
`R_i =  26 \Omega`

The second resistance is
`R_E =  200 \Omega`

Generally the equivalent resistance is

`R_e =  R_1 + R_E`

=>
`R_e =  26 +200`

=>
`R_e = 226 \ \Omega`

Generally the time constant is mathematically represented as

`\tau  =  RC`

=>
`\tau  =  226 * 150 *10^(-9)`

=>
`\tau  =  3.39 *10^(-5) \  s`

Generally the voltage is mathematically represented as

`V =  V_o  e^{-(t)/(\tau) }`

=>
`(V_o)/(2) =  V_o  e^{-(t)/(\tau) }`

=>
`0.5 =    e^{-(t)/(\tau) }`

=>
`ln(0.5) =    {-(t)/( 3.39 *10^(-5) ) }`

=>
`ln(0.5)   * 3.39 *10^(-5)  =   -t`

=>
`t = 2.35*10^(-5) \  s`